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\(y'=x^2-2\left(m-1\right)x+3\left(m-1\right)\)
Hàm đồng biến trên khoảng đã cho khi với mọi \(x>1\) ta luôn có:
\(g\left(x\right)=x^2-2\left(m-1\right)x+3\left(m-1\right)\ge0\)
\(\Rightarrow\min\limits_{x>1}g\left(x\right)\ge0\)
Do \(a=1>0;-\dfrac{b}{2a}=m-1\)
TH1: \(m-1\ge1\Rightarrow m\ge2\)
\(\Rightarrow g\left(x\right)_{min}=f\left(m-1\right)=\left(m-1\right)^2-2\left(m-1\right)^2+3\left(m-1\right)\ge0\)
\(\Rightarrow\left(m-1\right)\left(4-m\right)\ge0\Rightarrow1\le m\le4\Rightarrow2\le m\le4\)
TH2: \(m-1< 1\Rightarrow m< 2\Rightarrow g\left(x\right)_{min}=g\left(1\right)=m\ge0\)
Vậy \(0\le m\le4\)
a: \(y=-x^3+\left(m+2\right)x^2-3x\)
=>\(y'=-3x^2+2\left(m+2\right)x-3\)
=>\(y'=-3x^2+\left(2m+4\right)\cdot x-3\)
Để hàm số nghịch biến trên R thì \(y'< =0\forall x\)
=>\(\left\{{}\begin{matrix}\left(2m+4\right)^2-4\cdot\left(-3\right)\left(-3\right)< =0\\-3< 0\end{matrix}\right.\)
=>\(4m^2+16m+16-4\cdot9< =0\)
=>\(4m^2+16m-20< =0\)
=>\(m^2+4m-5< =0\)
=>\(\left(m+5\right)\left(m-1\right)< =0\)
TH1: \(\left\{{}\begin{matrix}m+5>=0\\m-1< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m>=-5\\m< =1\end{matrix}\right.\)
=>-5<=m<=1
TH2: \(\left\{{}\begin{matrix}m+5< =0\\m-1>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m>=1\\m< =-5\end{matrix}\right.\)
=>\(m\in\varnothing\)
b: \(y=x^3-3x^2+\left(1-m\right)x\)
=>\(y'=3x^2-3\cdot2x+1-m\)
=>\(y'=3x^2-6x+1-m\)
Để hàm số đồng biến trên R thì \(y'>=0\forall x\)
=>\(\left\{{}\begin{matrix}\text{Δ}< =0\\a>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3>0\\\left(-6\right)^2-4\cdot3\left(1-m\right)>=0\end{matrix}\right.\)
=>\(36-12\left(1-m\right)>=0\)
=>\(36-12+12m>=0\)
=>12m+24>=0
=>m+2>=0
=>m>=-2
a: \(y=-x^3-3x^2+\left(5-m\right)x\)
=>\(y'=-3x^2-3\cdot2x+5-m\)
=>\(y'=-3x^2-6x+5-m\)
Để hàm số nghịch biến trên R thì \(y'< =0\forall x\)
=>\(\left\{{}\begin{matrix}\text{Δ}< =0\\a< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(-6\right)^2-4\cdot\left(-3\right)\left(5-m\right)< =0\\-3< 0\end{matrix}\right.\)
=>\(36+12\left(5-m\right)< =0\)
=>\(36+60-12m< =0\)
=>\(-12m+96< =0\)
=>-12m<=-96
=>m>=8
b: \(y=x^3+\left(2m-2\right)\cdot x^2+mx\)
=>\(y'=3x^2+2\left(2m-2\right)\cdot x+m\)
=>\(y'=3x^2+\left(4m-4\right)x+m\)
Để hàm số đồng biến trên R thì y'>=0 với mọi x
=>\(\left\{{}\begin{matrix}\text{Δ}< =0\\a>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3>0\\\left(4m-4\right)^2-4\cdot3\cdot m< =0\end{matrix}\right.\)
=>\(16m^2-32m+16-12m< =0\)
=>\(16m^2-44m+16< =0\)
=>\(4m^2-11m+4< =0\)
=>\(\dfrac{11-\sqrt{57}}{8}< =m< =\dfrac{11+\sqrt{57}}{8}\)
a: \(y=-x^3-\left(m+1\right)x^2+3\left(m+1\right)x\)
=>\(y'=-3x^2-\left(m+1\right)\cdot2x+3\left(m+1\right)\)
=>\(y'=-3x^2+x\cdot\left(-2m-2\right)+\left(3m+3\right)\)
Để hàm số nghịch biến trên R thì \(y'< =0\forall x\)
=>\(\left\{{}\begin{matrix}\text{Δ}< =0\\a< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(-2m-2\right)^2-4\cdot\left(-3\right)\left(3m+3\right)< =0\\-3< 0\end{matrix}\right.\)
=>\(4m^2+8m+4+12\left(3m+3\right)< =0\)
=>\(4m^2+8m+4+36m+36< =0\)
=>\(4m^2+44m+40< =0\)
=>\(m^2+11m+10< =0\)
=>\(\left(m+1\right)\left(m+10\right)< =0\)
TH1: \(\left\{{}\begin{matrix}m+1>=0\\m+10< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m>=-1\\m< =-10\end{matrix}\right.\)
=>\(m\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}m+1< =0\\m+10>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< =-1\\m>=-10\end{matrix}\right.\)
=>-10<=m<=-1
b: \(y=-\dfrac{1}{3}x^3+mx^2-\left(2m+3\right)x\)
=>\(y'=-\dfrac{1}{3}\cdot3x^2+m\cdot2x-\left(2m+3\right)\)
=>\(y'=-x^2+2m\cdot x-\left(2m+3\right)\)
Để hàm số nghịch biến trên R thì \(y'< =0\forall x\)
=>\(\left\{{}\begin{matrix}\text{Δ}< =0\\a< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-1< 0\\\left(2m\right)^2-4\cdot\left(-1\right)\cdot\left(-2m-3\right)< =0\end{matrix}\right.\)
=>\(4m^2+4\left(-2m-3\right)< =0\)
=>\(m^2-2m-3< =0\)
=>(m-3)(m+1)<=0
TH1: \(\left\{{}\begin{matrix}m-3>=0\\m+1< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m>=3\\m< =-1\end{matrix}\right.\)
=>\(m\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}m-3< =0\\m+1>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< =3\\m>=-1\end{matrix}\right.\)
=>-1<=m<=3
a: \(y'< 0\)
=>\(\left(x-3\right)^3\cdot\left(x-1\right)^{22}\cdot\left(-3x-6\right)^7< 0\)
=>\(\left(x-3\right)\left(-3x-6\right)< 0\)
=>\(\left(x+2\right)\left(x-3\right)>0\)
=>\(\left[{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\)
y'>0
=>\(\left(x+2\right)\left(x-3\right)< 0\)
=>\(-2< x< 3\)
y'=0
=>\(\left[{}\begin{matrix}x-3=0\\x-1=0\\-3x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x=-2\end{matrix}\right.\)
Ta có bảng xét dấu sau:
| x | \(-\infty\) -2 1 3 +\(\infty\) |
| y' | - 0 + 0 + 0 - |
Vậy: Hàm số đồng biến trên các khoảng \(\left(-2;1\right);\left(1;3\right)\)
Hàm số nghịch biến trên các khoảng \(\left(-\infty;-2\right);\left(3;+\infty\right)\)
b: y'<0
=>\(\left(4x-3\right)^3\cdot\left(x^2-1\right)^{21}\left(3x-9\right)^7< 0\)
=>\(\left(4x-3\right)\left(3x-9\right)\left(x^2-1\right)< 0\)
=>\(\left(4x-3\right)\left(x-3\right)\left(x^2-1\right)< 0\)
TH1: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)>0\\x^2-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< \dfrac{3}{4}\end{matrix}\right.\\-1< x< 1\end{matrix}\right.\Leftrightarrow-1< x< \dfrac{3}{4}\)
TH2: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)< 0\\x^2-1>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{4}< x< 3\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow1< x< 3\)
y'>0
=>\(\left(4x-3\right)\left(x-3\right)\left(x^2-1\right)>0\)
TH1: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)>0\\x^2-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< \dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -1\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}\left(4x-3\right)\left(x-3\right)< 0\\x^2-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{4}< x< 3\\-1< x< 1\end{matrix}\right.\Leftrightarrow\dfrac{3}{4}< x< 1\)
Ta sẽ có bảng xét dấu sau đây:
| x | \(-\infty\) -1 3/4 1 3 +\(\infty\) |
| y' | + 0 - 0 + 0 - 0 + |
Vậy: Hàm số đồng biến trên các khoảng \(\left(-\infty;-1\right);\left(\dfrac{3}{4};1\right);\left(3;+\infty\right)\)
Hàm số nghịch biến trên các khoảng \(\left(-1;\dfrac{3}{4}\right);\left(1;3\right)\)
Nếu phương trình là \(\left(2m^2-5m+2\right)\left(x-1\right)^{2021}\left(x^{2020}-2\right)+2x^2-3=0\) thì còn có cơ hội giải quyết
Chứ đề đúng thế này thì e rằng không có cơ hội nào cả.
a/ \(y'=3mx^2-2\left(m+1\right)x+3m\)
Xet m=0 ko thoa man
Xet m khac 0
\(y'\ge0\Leftrightarrow\left(m+1\right)^2-9m^2\le0\Leftrightarrow8m^2-2m-1\ge0\)
\(\Leftrightarrow m^2+8\le0\left(vl\right)\) => ko ton tai m thoa man
b/ \(y'=mx^2-2mx+2m-1\)
m=0 ko thoa man
Xet m khac 0
\(y'\ge0\Leftrightarrow\left\{{}\begin{matrix}m>0\\m^2-m\left(2m-1\right)\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>0\\m^2-m\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>0\\\left[{}\begin{matrix}m\ge1\\m\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow m\ge1\)
\(f'\left(x\right)=4x^3-4x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Để \(g\left(x\right)_{min}>0\Rightarrow f\left(x\right)=0\) vô nghiệm trên đoạn đã cho
\(\Rightarrow\left[{}\begin{matrix}-m< -2\\-m>7\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m>2\\m< -7\end{matrix}\right.\)
\(g\left(0\right)=\left|m-1\right|\) ; \(g\left(1\right)=\left|m-2\right|\) ; \(g\left(2\right)=\left|m+7\right|\)
Khi đó \(g\left(x\right)_{min}=min\left\{g\left(0\right);g\left(1\right);g\left(2\right)\right\}=min\left\{\left|m-2\right|;\left|m+7\right|\right\}\)
TH1: \(g\left(x\right)_{min}=g\left(0\right)\Leftrightarrow\left\{{}\begin{matrix}\left|m-2\right|\le\left|m+7\right|\\\left|m-2\right|=2020\\\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m\ge\dfrac{5}{2}\\\left|m-2\right|=2020\end{matrix}\right.\) \(\Rightarrow m=2022\)
TH2: \(g\left(x\right)_{min}=g\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}\left|m+7\right|\le\left|m-2\right|\\\left|m+7\right|=2020\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m\le\dfrac{5}{2}\\\left|m+7\right|=2020\end{matrix}\right.\) \(\Rightarrow m=-2027\)